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Question
z1 and z2 are two complex numbers such that |z1| = |z2| and arg(z1) + arg(z2) = π, then show that z1 = `-barz_2`.
Solution
Let z1 = r1(cosθ1 + isinθ1)
And z2 = r2(cosθ2 + isinθ2) are polar form of two complex numbers z1 and z2.
Given that: |z1| = |z2|
⇒ r1 = r2 ......(i)
And arg(z1) + arg(z2) = π
⇒ θ1 + θ2 = π
⇒ θ1 = π – θ2
Now z1 = r1[cos(π – θ2) + isin(π – θ2)]
⇒ z1 = r1[–cosθ2 + isinθ2]
⇒ z1 = –r1(cosθ2 – isinθ2) ......(i)
z2 = r2[cosθ2 + isinθ2]
`barz_2` = r1[cosθ2 – isinθ2] ......[∵ r1 = r2] .....(ii)
From equation (i) and (ii) we get,
z1 = `-barz_2`.
Hence proved.
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